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Ivanshal [37]
2 years ago
6

If h(x) = x - 7 and g(x) = x2, which expression is equivalent to (g0h)(5)?

Mathematics
1 answer:
notsponge [240]2 years ago
6 0

Answer:

A

Step-by-step explanation:

I honestly couldn't tell you how to do this, I don't understand it. I just took the test and got this question correct. The answer is A, (5-7)^2.

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A computer system requires that passwords contain at least one digit. if eight characters are generated at random, and each is e
KIM [24]
Figure this out as though no digits are present. You can choose 8 of the 26 letters and that's it.
26^8 [you can allow repeating letters]  =  2.088 * 10^11
The total number of ways that can be chosen with letters and digits intermixed = 36^8

The probability of getting no numbers is 26^8 / 36^8 or (26/36)^8 = 0.07402
So the probability that there were be at least one digit is
1 - 0.07402 = 0.92598
6 0
3 years ago
Help... anyone i just need help with these questions!?!
anyanavicka [17]
For the first one it is
4 0
3 years ago
Read 2 more answers
NEED ASAP
Gala2k [10]

Answer:

(- 2, 4 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) , then the midpoint is

( \frac{x_{1}+x_{2}  }{2} , \frac{y_{1}+y_{2}  }{2} )

Here (x₁, y₁ ) = A (4, 6 ) and (x₂, y₂ ) = B (- 8, 2 )

midpoint = ( \frac{4-8}{2} , \frac{6+2}{2} ) = ( \frac{-4}{2} , \frac{8}{2} ) = (- 2, 4 )

6 0
2 years ago
Read 2 more answers
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
3 years ago
You buy 1.86 pounds of ground beef 2,8
Mariana [72]
What is this question asking?
maybe try 3.87
7 0
3 years ago
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